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I have these two questions that I cannot get any intuition about. Perhaps someone can possibly offer a few hints on how to get started?

1) Show that the ends of ${\bf F_2} \oplus {\bf F_2}$ is equal to 1 (i.e, $e({\bf F_2} \oplus {\bf F_2}) =1$). The reason this is confusing me is because $e({\bf F_2}) = \infty$, so I do not know how my question is equal to 1.

2) Let $G$ be a finitely generated group with $e(G)=2$ and $\lambda$ be its cayley graph. There exists a finite subgraph $C$ such that $\lambda$ \ $C$ has exactly two connected, unbounded components. We choose one of these two complements, and now let $E \subset G$ consist of the elements of $G$ that correspond to the vertices in a connected component that is unbounded. Note that $g \in G$. I need to show that: either $E \triangle gE$ is finite or $(E \triangle gE)^{c}$ is finite.

Any help would be great.

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1 Answer 1

Let me give you a rough idea of how to prove 1). Presumably ${\bf F_2}$ is the free group of rank 2? We have to show that two points in the Cayley graph of ${\bf F_2} \oplus {\bf F_2}$ at distance at least $n$ from the origin can be connected by a path that stays at distance at least $n$ from the origin. Let the two points be $(g_1,h_1)$, $(g_2,h_2)$, and let $g$ be any element of ${\bf F_2}$ that has length at least $n$. Then we use the path

$(g_1,h_1) \to (g,h_1) \to (g,h_2) \to (g_2,h_2)$,

where, in each component of the path, one of the co-ordinate elements stays constant.

I don't think you have stated 2) very clearly. You introduced a symbol $\lambda$ and then never used it. Connected component of what? You cannot mean $\lambda$, because that is connected!

Added later: Hints for 2). Let the two unbounded connected components of $\lambda \setminus C$ be $E=E_1,E_2$, where $\lambda \setminus(E_1 \cup E_2)$ is finite. Now $\lambda \setminus (gC \cup g^{-1}C)$ also has exactly two infinite unbounded connected components $F_1,F_2$ with $\lambda \setminus(F_1 \cup F_2)$ finite, and we can assume that $F_1 \subset E_1$, $F_2 \subset E_2$.

Now any two points in $gF_1$ are in the same connected component of $\lambda \setminus C$, and similarly for $gF_2$. So we must either have $gF_1 \subset E_1$, $gF_2 \subset E_2$, or $gF_1 \subset E_2$, $gF_2 \subset E_1$.

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@Derek..I added some more info about $E$. Does this look better? –  Selena Jun 11 '12 at 1:02

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